Question

At time t the resultant force on a particle, of mass 250kg is (300ti-400tj)N. Initially, the particle is at the origin and is moving
with the velocity (2i-3j)m
−1
.
a) Find the acceleration at time t.
b) Find the velocity of the particle at time t
c) Find the position vector of the particle at time t

Answer 1

Answer:

At time t the resultant force on a particle, of mass 250kg is (300ti-400tj)N. Initially, the particle is at the origin and is moving

with the velocity (2i-3j)m

−1

Answer 2

a) Acceleration is $\dfrac{6}{5}t \hat{i} - \dfrac{8}{5}t \hat{j} \ m/s^2$

b) Velocity is $\dfrac{3t^2 + 10}{5} \hat{i} - \dfrac{4t^2 + 15}{5} \hat{j} \ m/s$

c) Position Vector is $\dfrac{3t^{3} + 30t }{15} \hat{i} - \dfrac{4t^{3} + 45}{15} \hat{j} \ m$

Step-by-step explanation:

Given: Resultant force on a particle = $\vec{F} = 300t \hat{i} - 400t \hat{j} \ N$

Mass of the particle = $m = 250 \text{ kg}$

Velocity of the particle at the origin = $\vec{u} = 2 \hat{i} - 3 \hat{j} \ m$

To Find: a) acceleration, b) velocity and c) position vector of the particle at time 't'

Solution:

• Finding acceleration of the particle at time 't'

The force $\vec{F}$ is equal to the product of the mass $'m'$ and the acceleration $' \vec{a} \ '$ of the particle $\Rightarrow \vec{F} = m \vec{a}$

Therefore, the acceleration will be,

$\Rightarrow \text{acceleration, } \vec{a} = \dfrac{\vec{F}}{m}$

$\Rightarrow \text{acceleration, } \vec{a} = \dfrac{300t \hat{i} - 400t \hat{j} }{250} = \dfrac{300}{250}t \hat{i} - \dfrac{400}{250}t \hat{j}$

$\Rightarrow \text{acceleration, } \vec{a} = \dfrac{6}{5}t \hat{i} - \dfrac{8}{5}t \hat{j} \ m/s^2$

• Finding velocity of the particle at time 't'

The relation between the acceleration $' \vec{a} \ '$ and velocity $' \vec{v} \ '$ of the particle is  $\Rightarrow \vec{v} = \int \vec{a} dt + \vec{u}$ given $' \vec{u} \ '$ is the velocity of the particle at the origin.

Therefore, the velocity of the particle will be,

$\Rightarrow \text{velocity, } \vec{v} = \int \vec{a} dt + \vec{u}$

$\Rightarrow \text{velocity, } \vec{v} = \int (\dfrac{6}{5}t \hat{i} - \dfrac{8}{5}t \hat{j}) dt + (2\hat{i} - 3\hat{j})$

$\Rightarrow \text{velocity, } \vec{v} =\dfrac{3}{5}t^2 \hat{i} - \dfrac{4}{5}t^2 \hat{j} + 2\hat{i} - 3\hat{j}$

$\Rightarrow \text{velocity, } \vec{v} = \dfrac{3t^2 + 10}{5} \hat{i} - \dfrac{4t^2 + 15}{5} \hat{j} \ m/s$

• Finding position vector of the particle at time 't'

For the velocity $' \vec{v} \ '$ of the particle, the position vector is  $\Rightarrow \vec{r} = \int \vec{v} dt$

Therefore, the position vector of the particle will be,

$\Rightarrow \text{position vector, } \vec{r} = \int \vec{v} dt$

$\Rightarrow \text{position vector, } \vec{r} = \int ( \dfrac{3t^2 + 10}{5} \hat{i} - \dfrac{4t^2 + 15}{5} \hat{j}) dt$

$\Rightarrow \text{position vector, } \vec{r} = \dfrac{3t^3 + 30}{15} \hat{i} - \dfrac{4t^3 + 45}{15} \hat{j}$

Hence, a) Acceleration is $\dfrac{6}{5}t \hat{i} - \dfrac{8}{5}t \hat{j} \ m/s^2$

b) Velocity is $\dfrac{3t^2 + 10}{5} \hat{i} - \dfrac{4t^2 + 15}{5} \hat{j} \ m/s$

c) Position Vector is $\dfrac{3t^{3} + 30t }{15} \hat{i} - \dfrac{4t^{3} + 45}{15} \hat{j} \ m$